| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Deduce related integral from numerical approximation |
| Difficulty | Moderate -0.3 Part (a) is a routine trapezium rule application with given ordinates requiring only formula substitution and arithmetic. Part (b) requires recognizing the relationship between the shaded area and the given integral (likely involving a transformation or area interpretation), which adds mild problem-solving but remains a standard exam technique. Overall slightly easier than average due to the mechanical nature of trapezium rule and the deduction being a common question type. |
| Spec | 1.08d Evaluate definite integrals: between limits1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(h = 0.6\) OE; accept 0.3 OE as the multiplier | B1 | PI by correct answer |
| Substitutes given \(y\) values: \(2.90988 + 0.09329 + 2(1.26485 + 0.62305 + 0.32374 + 0.17263)\) | M1 | Ignore \(h\); accept correct exact values or values to more than 5 decimal places; PI by correct answer or 7.77171 |
| \(\frac{0.6}{2}\Big(2.90988 + 0.09329 + 2(1.26485 + 0.62305 + 0.32374 + 0.17263)\Big) = 2.3315\) | A1 | Obtains 2.3315 AWRT |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(9.3\) | R1F | Obtains \(4\times\) their answer to (a), correct to at least 2 significant figures |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $h = 0.6$ OE; accept 0.3 OE as the multiplier | B1 | PI by correct answer |
| Substitutes given $y$ values: $2.90988 + 0.09329 + 2(1.26485 + 0.62305 + 0.32374 + 0.17263)$ | M1 | Ignore $h$; accept correct exact values or values to more than 5 decimal places; PI by correct answer or 7.77171 |
| $\frac{0.6}{2}\Big(2.90988 + 0.09329 + 2(1.26485 + 0.62305 + 0.32374 + 0.17263)\Big) = 2.3315$ | A1 | Obtains 2.3315 AWRT |
## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $9.3$ | R1F | Obtains $4\times$ their answer to (a), correct to at least 2 significant figures |
---5
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with 6 ordinates ( 5 strips) to find an approximate value for the shaded area.
Give your answer to four decimal places.\\
5
\item Using your answer to part (a) deduce an estimate for $\int _ { 1 } ^ { 4 } \frac { 20 } { \mathrm { e } ^ { x } - 1 } \mathrm {~d} x$
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2023 Q5 [4]}}